[LeetCode]Subsets II-白红宇

[LeetCode]Subsets II

阅读量：4150 次

发布时间：2019-05-25

本文共 1986 字，大约阅读时间需要 6 分钟。

class Solution {//because Elements in a subset must be in non-descending order.//so if we sort the vector first.//If we choose an elem[i] into subset, then we should choose elem[j](i

& S, int curPos, vector

& oneSubset, vector

>& allSubset) { allSubset.push_back(oneSubset); for (int i = curPos; i < S.size(); ++i) { if(i != curPos && S[i] == S[i-1]) continue; oneSubset.push_back(S[i]); DFS(S, i+1, oneSubset, allSubset); oneSubset.pop_back(); } }public: vector

> subsetsWithDup(vector

&S) { // Start typing your C/C++ solution below // DO NOT write int main() function sort(S.begin(), S.end()); vector

oneSubset; vector

> allSubset; DFS(S, 0, oneSubset, allSubset); return allSubset; }};

second time

class Solution {public:    void subsetUtil(vector
   
    & S, vector
    
     & used, int curIdx, vector
     
      & curPath, vector
      
       
         >& allPath)    {        if(curIdx == S.size())        {            allPath.push_back(curPath);            return ;        }                subsetUtil(S, used, curIdx+1, curPath, allPath);        if(curIdx >= 1 && S[curIdx] == S[curIdx-1] && used[curIdx-1] == false) return;        used[curIdx] = true;        curPath.push_back(S[curIdx]);        subsetUtil(S, used, curIdx+1, curPath, allPath);        curPath.pop_back();        used[curIdx] = false;    }    vector
        
         
           > subsetsWithDup(vector
          
            &S) { // Start typing your C/C++ solution below // DO NOT write int main() function sort(S.begin(), S.end()); vector
           
             used(S.size(), false); vector
            
             
               > allPath; vector
              
                curPath; subsetUtil(S, used, 0, curPath, allPath); return allPath; }};